Answer
$\mathscr{E} = 0.30~V$
Work Step by Step
$i = \frac{\mathscr{E}}{R+r}$
We can find an expression for $\mathscr{E}$:
$V = \mathscr{E}-ir$
$V = \mathscr{E}-\frac{\mathscr{E}~r}{R+r}$
$V = \frac{\mathscr{E}~R}{R+r}$
$\mathscr{E} = \frac{V~(R+r)}{R}$
We can use the given values to find the internal resistance $r$:
$\mathscr{E} = \frac{(0.10)~(500+r)}{500} = \frac{(0.15)(1000+r)}{1000}$
$(2)(0.10)~(500+r) = (0.15)(1000+r)$
$0.20~r-0.15~r = 150-100$
$0.05~r = 50$
$r = \frac{50}{0.05}$
$r = 1000~\Omega$
The internal resistance is $~~1000~\Omega$
We can find the emf of the solar cell:
$\mathscr{E} = \frac{V~(R+r)}{R}$
$\mathscr{E} = \frac{(0.10~V)~(500~\Omega+1000~\Omega)}{500~\Omega}$
$\mathscr{E} = 0.30~V$
