Answer
(a) $v = 0.9998~c$
(b) $t = 1.0\times 10^{-11}~s$
Work Step by Step
(a) We can find the speed:
$K \approx \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$
$\sqrt{1-\frac{v^2}{c^2}} = \frac{mc^2}{K}$
$1-\frac{v^2}{c^2} = (\frac{mc^2}{K})^2$
$\frac{v^2}{c^2} = 1-(\frac{mc^2}{K})^2$
$\frac{v^2}{c^2} = 1-(\frac{(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2}{25\times 10^6~eV})^2$
$\frac{v^2}{c^2} = 1-(\frac{(8.19\times 10^{-14}~N)(6.24\times 10^{18}~eV/N)}{25\times 10^6~eV})^2$
$\frac{v^2}{c^2} = 1-(\frac{5.11\times 10^5~eV}{25\times 10^6~eV})^2$
$\frac{v^2}{c^2} = 0.99958$
$v = \sqrt{0.99958}~c$
$v = 0.9998~c$
(b) We can find $L$, the distance in the electron frame:
$L = L_0~\sqrt{1-\frac{v^2}{c^2}}$
$L = (15~cm)~\sqrt{1-\frac{(0.9998~c)^2}{c^2}}$
$L = (15~cm)~\sqrt{1-(0.9998)^2}$
$L =0.30~cm$
We can find the time it takes the electron to move this distance:
$t = \frac{3.0\times 10^{-3}~m}{(0.9998)(3.0\times 10^8~m/s)}$
$t = 1.0\times 10^{-11}~s$
