jmc

The water in the tank fills a cone, which we will refer to as the water cone, that is similar to the cone-shaped tank itself. Let the scale factor between the water cone and tank be $x$, so the height of the water cone is $96x$ feet and the radius of the water cone is $16x$ feet. It follows that the volume of the water cone is $(1/3)\pi (16x{)}^2(96x)$ cubic feet.

The volume of the cone-shaped tank is $(1/3)\pi (16^2)(96)$. Since the water cone has $25\%$ or 1/4 of the volume of the tank, we have $$(1/3)\pi (16x{)}^2(96x)=(1/4)(1/3)\pi (16^2)(96).$$ Simplifying yields $x^3=1/4$, so $x=\sqrt[3]{1/4}$.

Finally, the height of the water in the tank is the height of the water cone, which is $$96x=96\sqrt[3]{1/4}=48\cdot 2\sqrt[3]{1/4}=48\sqrt[3]{(1/4)(8)}=48\sqrt[3]{2}$$ feet. Therefore, we have $a+b=48+2=\boxed{50}$.