The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is

Given that, The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm
So, $$2(lb+bh+hl)=846$$. 

And $$4(l+b+h)=144$$
$$(l+b+h)=36$$
$$\left(l+b+h\right)^2=l^2+b^2+h^2+2\left(lb+bh+hl\right)$$
$$1296=\left(l^2+b^2+h^2\right)+846$$
$$450=l^2+b^2+h^2$$

We are told that this cuboid is inscribed in a sphere, the body diagonal of the cuboid equals the diameter of the sphere, this can be visualised as:

This is nothing but, $$\sqrt{l^2+b^2+h^2}=2R$$
$$l^2+b^2+h^2=4R^2$$
$$450=4R^2$$
$$R^2=\frac{225}{2}$$
$$R=\frac{15}{\sqrt{2}}$$

Volume of sphere will be $$\dfrac{4}{3}\times\ \pi\ \times\ \left(\dfrac{15}{\sqrt{2}}\right)^3$$

$$\dfrac{4}{3}\pi\ \left(\dfrac{3375}{2\sqrt{\ 2}}\right)$$

$$\pi\ \times\ 1125\sqrt{\ 2}$$