#### Answer

(a) The kayak should be directed at an angle of $41.8^{\circ}$ west of north.
(b) It will take 45 seconds to cross the harbor.

#### Work Step by Step

(a) The current will push the kayak to the east at a speed of 2.0 m/s. To go directly north, a component of the kayak's velocity should be directed to the west at 2.0 m/s. Let $\theta$ be the angle of the kayak's velocity vector which is west of north.
$v_{west} = v~sin(\theta) = 2.0~m/s$
$sin(\theta) = \frac{2.0~m/s}{v}$
$\theta = arcsin(\frac{2.0~m/s}{3.0~m/s})$
$\theta = 41.8^{\circ}$
The kayak should be directed at an angle of $41.8^{\circ}$ west of north.
(b) We can find the north component of the kayak's velocity vector.
$v_{north} = (3.0~m/s)~cos(41.8^{\circ})$
$v_{north} = 2.24~m/s$
We can find the time to cross the harbor.
$t = \frac{d}{v_{north}}$
$t = \frac{100~m}{2.24~m/s}$
$t = 45~s$
It will take 45 seconds to cross the harbor.