I want answer with full solution as soon as possible exams tomorrow plz !!!!!
Have posted question in image attachment,

To determine the centre of mass of the body shown above, we will consider this body as a combinition of two spheres, one is sphere of radius 'R' and mass 'M', and the another one of radius 'a' and mass '-m', also the mass of this sphere will be assumed negative, so that the sum of the masses of the both of the spheres becomes equal to the mass of the given body. 
       Let the density of the body be '$\rho$'.
(1). Mass of bigger sphere:
                                        $M = \frac{4}{3}\pi R^3\rho$                          .....eq(1)
(2). Mass of smaller sphere:
                                      $m = \frac{4}{3}\pi a^3\rho$                           ......eq(2)
Now we will calculte the C.O.M. of the given body
    Formula:                  $r_{CM} = \frac{m_1{r}_1+m_2{r}_2}{m_1+m_2}$
(1). 'X' co-ordinate of C.O.M.:
                                   $$\begin{gathered} X_{CM} = \frac{(M\times 0)+(-m\times b)}{\left(M\right)+(-m)} \\ {X}_{CM} = \frac{({\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\pi R^3\rho \times 0)+(-{\displaystyle \frac{4}{3}}\pi a^3\rho \times b)}{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\pi R^3\rho -\frac{4}{3}\pi a^3\rho } \\ {X}_{CM} = \frac{-a^{3}b}{R^3-a^3} \end{gathered}$$
        
(2). 'Y' co-ordinate of C.O.M.:
                                  
                                   $$\begin{gathered} Y_{CM} = \frac{(M\times 0)+(-m\times 0)}{\left(M\right)+(-m)} \\ {Y}_{CM} = 0 \end{gathered}$$
(3). 'Z' co-ordinate of C.O.M.
                                 $$\begin{gathered} Z_{CM} = \frac{(M\times 0)+(-m\times 0)}{\left(M\right)+(-m)} \\ {Z}_{CM} = 0 \end{gathered}$$
Ans: $[-\frac{a^{3}b}{(R^3-a^3)},0,0]$