Answer
The width according to the proton's reference frame is $~~2.21~mm$
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 $~~2.21~mm$
