Answer
12.0 seconds
Work Step by Step
Using the law of cosines, $\big(\sqrt{(R+0.2m)^2-R^2\big)^2}=(R+0.2)^2+R^2-2\times R\times (R+0.2)\times \cos(x)$
Simplifying, we get $\cos(x)=\frac{R}{R+0.2}$
The earth's radius is 6.38$\times10^6m$ and plugging this into the equation, we get $x=0.0143$
Similarly, we get $cos(y)=\frac{R}{R+1.5}$ and $y=0.0393$
The angle the sun moved can be calculated using $\theta=2y-2x=0.0499^o$
The time can be calculated by a series of conversion factors.
$t =0.0499^o\times \frac{1 day}{360^o} \times \frac{24 hr}{1 day} \times \frac{60 min}{1 hr} \times \frac{60 s}{ 1 min}=12.0s$
