Explanation

(a) Here, Magnetic field, $B = 1.2mT = 1.2 \times 10^{- 3}T$
Mass of the proton, $m_{p} = 1.67 \times 10^{- 27}kg$
Speed of the proton enters a uniform magnetic field perpendicular to its velocity, it follows a circular path and whose radius is given by
$R = \frac{m_{p}v}{qB}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Centripetal acceleration of the proton is $a_{c} = \frac{v^{2}}{R}$
$= \frac{v^{2}}{\left( \frac{m_{p}v}{qB} \right)} = \frac{vqB}{m_{p}}\ \ \ \ \ \ \ \ \ \lbrack Using\ (i)\rbrack$
$= \frac{(3 \times 10^{7}ms^{- 1}) \times (1.6 \times 10^{- 19}C) \times (1.2 \times 10^{- 3}T)}{(1.67 \times 10^{- 27}kg)}$
$= 3.45 \times 10^{12}ms^{- 2}$