A solid spherical region has a spherical cavity having a diameter R (equal to the radius of the spherical region), has a total charge Q . The volumetric charge density is $\rho$. The electric field and potential at a point P at x = 2R as shown is 

Charge Density $\rho =\frac{Q}{\frac{4}{3}\pi (R^3-{\left(\frac{R}{2}\right)}^3)}=\frac{6Q}{7\pi R^3}$

Using the Superposition Principle,

$V=\frac{\rho \left(\frac{4}{3}\pi R^3\right)}{4\pi {\epsilon }_{0}x}-\frac{\rho \left(\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3\right)}{4\pi {\epsilon }_0(x-\frac{R}{2})}=\frac{\rho R^3}{3{\epsilon }_0}\frac{(7x-4R)}{4x(2x-R)}$

At x=2R, we get $V=\frac{5}{36}\left(\frac{\rho R^2}{{\epsilon }_0}\right)$

Similarly,$E=\frac{\rho \left(\frac{4}{3}\pi R^3\right)}{4\pi {\epsilon }_0{x}^2}-\frac{\rho \left(\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3\right)}{4\pi {\epsilon }_0{(x-\frac{R}{2})}^2}$

$E=\frac{\rho R^3}{3{\epsilon }_0}\left(\frac{1}{x^2}-\frac{1}{2{(2x-R)}^2}\right)$

So, at x = 2R, we get $E=\frac{7\rho R}{108{\epsilon }_0}=\frac{7}{108}\left(\frac{\rho R}{{\epsilon }_0}\right)$