Answer
$\frac{q_2}{q_1}=1.125$.
Work Step by Step
Applying Formula, we have
$
E_1=\frac{\left|q_1\right|}{4 \pi \varepsilon_0 R^3} r_1=\frac{\left|q_1\right|}{4 \pi \varepsilon_0 R^3}\left(\frac{R}{2}\right)=\frac{1}{2} \frac{\left|q_1\right|}{4 \pi \varepsilon_0 R^2} .
$
Also, outside sphere 2 we have
$
E_2=\frac{\left|q_2\right|}{4 \pi \varepsilon_0 r^2}=\frac{\left|q_2\right|}{4 \pi \varepsilon_0(1.50 R)^2} .
$
Equating these and solving for the ratio of charges, we arrive at $\frac{q_2}{q_1}=\frac{9}{8}=1.125$.
