Only differentiation. eeekkkk!

Ok. Here's what I'd do. I'm not at all confident but there's nothing else I can suggest.

Let's assume the sides of the block are always given by those expressions in x whatever size x is.

Then you can write

hence work out dV/dx using differentiation.

You can also write an expression for the surface area in terms of x and hence write the rate of change of volume

You should be able to work out both numbers.

Stick a minus on this as the volume is decreasing.

As x squared occurs in both of these you can get

This will integrate to an expression like this

m will be negative. Note x will be the vertical axis and t across.

Use the initial volume and volume after one unit ot time to determine this gradient m.

When does the line cut x = 0 ?

That's my best shot I'm afraid. Maybe someone else on the forum will come in with a better suggestion.

Bob