The relation between the volume $ V $ and the mass $ M $ of a nucleus is
(A) $ V \propto {M^3} $
(B) $ V \propto {M^{1/3}} $
(C) $ V \propto M $
(D) $ V \propto \dfrac{1}{M} $

Hint: To answer this question, we need to use the relationship of the radius of the nucleus with the mass number. Then using the relation of the mass with the volume, i.e. the density, we can get the required proportionality.

Formula used:
 $ R = {R_0}{A^{1/3}} $ , here $ R $ is the radius of the nucleus, $ A $ is the mass number of the nucleus, and $ {R_0} $ is a constant.
 $ D = \dfrac{M}{V} $ , here $ D $ is the density of the nucleus, $ M $ is the mass of the nucleus, and $ V $ is the volume of the nucleus.

Complete step by step solution:
The relationship between the radius and the mass number of a nucleus has been found experimentally to be
 $ R = {R_0}{A^{1/3}} $ .....................(1)
We know that the volume of a nucleus, $ V \propto {R^3} $
From (1), we can say that the volume $ V \propto A $
Also, we know that the mass of a nucleus is proportional to the mass number $ A $ .
That is, $ M \propto A $
Now, we know that the density $ D = \dfrac{M}{V} $ .....................(2)
As both the mass and the volume are proportional to the mass number $ A $ , the density becomes independent of $ A $ , or it is a constant.
So, from (2) we have
 $ D = \dfrac{M}{V} $
On cross-multiplying we get
 $ V = \dfrac{M}{D} $
As $ D $ is a constant, so $ \dfrac{1}{D} $ is also a constant.
 $ \therefore V \propto M $
So, the volume of a nucleus is proportional to its mass.

Note:
The above proportionality can also be obtained without even considering the relation of the individual quantities with the mass number. As we know that the density of the nuclear matter is a constant ( $ \approx 2.3 \times {10^{23}}kg{m^{ - 3}} $ ), so the volume of the nucleus will be linearly proportional to its mass.