Say you fire a 10kg steel round at the incoming tungsten round at a comparatively paltry 3 km/s
This is no different to having a static steel Whipple shield, and an incoming projectile travelling at 63km/s. The problem is still that hypervelocity impacts do not behave like low velocity ones, and as a result the effects of collisions are counterintuitive. This is why more sedate collisions (say, 15km/s) are modelled as jets of fluid splashing off each other, for example... intermolecular bonds don't mean anything given the impact forces involved.
This means that all the blurb below is just as applicable to whipple shields as it is to interceptor projectiles.
Will the tungsten round be vaporized? Will it be blown to fist-sized chunks?
It isn't entirely clear what will happen to it. Worst case scenario is that a short length of the impactor is ablated off, and the rest just carried on coming. Using the same hydrodynamic penetration approximation as I did last time... if your steel projectile has the same diameter as the incoming projectile it'll be 4cm thick, and ablated about 2.5cm off the front of the impact. That's not so good, from the target's point of view.
I've recently been reading up on "crater strength", a notion that handles the expansion of a crater in a solid object. I'm not entirely certain how this applies to hypervelocity impactors... it doesn't gel nicely with the idea of hydrodynamic jets, that's for sure, but does handle the notion of something exploding, as you might expect releasing an awful lot of energy in a short period of time. It was also suggested by Luke Campbell, who knows a fair bit more about this sort of thing than I do, and has put more thought into it. So with all that said, take this with a small pinch of salt.
The cratering approximation define crater volume $V_c = E_p/S_c$ where $E_p$ is the kinetic energy of the projectile and $S_c$ is the cratering strength of the material involved, handwaved to be three times its yield strength. The yield strength of tungsten is 750MPa, so its cratering strength is defined as 2.25GJ/m3. If we imaging the tungsten impactor as stationary, and the steel interceptor is coming in at 63km/s, it'll have a kinetic energy of nearly 20GJ. That gives a cratering volume of 8.82m3, and hence a cratering depth (as defined as the radius of a sphere with that crater volume) of about 2.1m.
With that approximation, the impactor is indeed blown to pieces. Hooray! (it also suggests your whipple shielding in the previous question is better than originally anticipated, so I'll revisit that at some point).
Given the hydrodynamic penetration depth assumption, the steel interceptor will be more or less "used up" in the front couple of centimetres of the impactor. The energy of the collision must, therefore, be transmitted along the impactor by plain old atoms bumping together. If that happens at the speed of sound in tungsten, 5.2km/s, it'll take 1/5200th of a second for the impactor to fully disintegrate, during which time it will have travelled 11m. If the interceptor hits the impactor any closer to the ship than that, you're still in big trouble (this also suggests your 50m whipple shielding gap in your previous answer is probably a sensible spacing). I'm not sure what speed the debris will expand at, as that requires working out energy budgets and things and I'm feeling too lazy for that. The speed at which the projectile expands should inform you as to how far away you must intercept the projectile, and how much armour you'll need to mop up the bits.
Next, there's a lot of momentum in a quarter-tonne slug travelling at 60km/s. Your little steel projectile may deliver enough energy to break it up, but those bits are going to be quite big and will maintain most of their original speed and direction. This isn't so much "spalling" as a terrifying shotgun blast of doom. Not as much doom as a metre long rod of tungsten, but you might still need to deal with a cloud of fist-sized 60km/s projectiles (it won't matter if they're solid or melted, incidentally).
A cloud of plasma can be dealt with by a magnetic shield; a cloud of dust can be dealt with by a whipple shield. Fist-sized chunks? You need some pretty hefty armor. Mostly-intact round? Good luck.
I don't think it will be dust. And remember, even if it was dust, a quarter tonne of dust travelling at 60km/s requires a pretty substantial whipple shield, and that whipple shield will have a pretty substantial hole in it afterwards.
As for dealing with plasma via a magnetic shield... again, the remnants of the impactor will have considerable momentum, and you have a short period of time in which to exert a considerable force on it in order to deflect it. I'm not going to try and work this out here (magnet maths is hard :-() but it sounds a little dubious.
Finally though, let's revisit those fist sized chunks. At the end of my last answer, I said this:
the simplest countermeasure from the attacker's point of view is to fire multiple smaller projectiles, slightly separated along their trajectory.
This still holds true. Colinear kinetic penetrators will be excellent at defeating defense in depth, though sufficient interceptor railgun fire might be able to swat all the projectiles before they reach the target. The other alternative, many long, slender projectiles (a sort of parallel impactor instead of serial) would require one railgun round each, which may very very quickly overwhelm any plausible defense. A combination of the two approaches sacrifices the sheer one-hit-kill ability of the monolithic tungsten slug for a huge cloud of kinetic death which will be impractical to defend against.