Answer
$E = 7.31~N/C$
Work Step by Step
We can find the charge in the spherical shell between radius $r = a$ and $r = 1.50a$:
$q = \rho~(\frac{4}{3}\pi~(1.50a)^3-\frac{4}{3}\pi~a^3)$
$q = (\rho)~(\frac{4}{3}\pi)~[(1.50a)^3-a^3]$
$q = (1.84\times 10^{-9}~C/m^3)~(\frac{4}{3}\pi)~[(0.150~m)^3-(0.100~m)^3]$
$q = 1.83\times 10^{-11}~C$
We can draw a Gaussian sphere with radius $r = 1.50a$ that is concentric with the spherical shell.
We can find the magnitude of the electric field at $r = 1.50a$:
$\epsilon_0~\Phi = q$
$(\epsilon_0)(E)(4\pi~r^2) = q$
$E = \frac{q}{4\pi~\epsilon_0~r^2}$
$E = \frac{1.83\times 10^{-11}~C}{(4\pi)~(8.854\times 10^{-12}~F/m)~(0.150~m)^2}$
$E = 7.31~N/C$
