The volume \(V\) of a sphere is given by \(V = \frac{4}{3}\pi r^3\).

Differentiate with respect to time \(t\):

\(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\).

We know \(\frac{dV}{dt} = 50\) cm³s^-1 and \(r = 10\) cm.

Substitute these values into the differentiated equation:

\(50 = 4\pi (10)^2 \frac{dr}{dt}\).

Simplify:

\(50 = 400\pi \frac{dr}{dt}\).

Solve for \(\frac{dr}{dt}\):

\(\frac{dr}{dt} = \frac{50}{400\pi} = \frac{1}{8\pi}\).

Thus, the rate at which the radius is increasing is \(\frac{1}{8\pi}\) cm/s or approximately 0.0398 cm/s.