From the Napier's circle (SIN-TAAD rule for a)

$\sin a = \tan b \tan \bar{D}$

$\sin 14^\circ = \tan b \tan (90^\circ - 30^\circ)$

$\tan b = \dfrac{\sin 14^\circ}{\tan 60^\circ}$

$b = 7^\circ 57'$

From triangle NAC (Cosine Law for Sides)

$\cos b = \cos 90^\circ \cos 90^\circ + \sin 90^\circ \sin 90^\circ \cos \alpha$

$\cos 7^\circ 57' = 0(0) + 1(1) \cos \alpha$

$\alpha = 7^\circ 57'$

$\theta = 121.5^\circ - \alpha$

$\theta = 121.5^\circ - 7^\circ 57'$

$\theta = 113.55^\circ$

Thus, the airplane will pass the equator at 113° 33' East of the prime meridian. *answer: C*