Answer
a) 3.7 fm
b) 240
c) 240
Work Step by Step
(a) We know that
$r=(1.2\times 10^{-15})A^{\frac{1}{3}}$
$\implies r=(1.2fm)(30)^{\frac{1}{3}}$
$r=3.7fm$
(b) We can find the required mass number as follows:
$(2r)^3=8r^3$
Now $A=8(30)=240$
(c) As $r=(1.2fm)A^{\frac{1}{3}}$
$r=2r_p$
$\implies r=2(3.7fm)=7.4fm$
Now $A^{\frac{1}{3}}=\frac{7.4fm}{1.2fm}=6.2$
$A=(6.2)^3=240$