Answer
$t = 7.37\times 10^{-12}~s$
Work Step by Step
We can find $\gamma$:
$E = \gamma mc^2 = (14.24~nJ)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$(\gamma) (938~MeV) = 89~GeV$
$\gamma = \frac{89~GeV}{938~MeV}$
$\gamma = 94.88$
We can find the width according to the proton's reference frame:
$L = \frac{L_0}{\gamma}$
$L = \frac{0.210~m}{94.88}$
$L = 0.00221~m$
$L = 2.21~mm$
The width according to the proton's reference frame is $~~2.21~mm$
We can find $\beta$:
$\gamma = \frac{1}{\sqrt{1-\beta^2}}$
$1-\beta^2 = \frac{1}{\gamma^2}$
$\beta^2 = 1-\frac{1}{\gamma^2}$
$\beta = \sqrt{1-\frac{1}{\gamma^2}}$
$\beta = \sqrt{1-\frac{1}{(94.88)^2}}$
$\beta = 0.9999445$
We can find the time according to the proton's frame:
$t = \frac{2.21\times 10^{-3}~m}{0.9999445~c}$
$t = \frac{2.21\times 10^{-3}~m}{(0.9999445)(3.0\times 10^8~m/s)}$
$t = 7.37\times 10^{-12}~s$
