The way the question is phrased, it sounds like the height stays fixed at 18 in. If L and W denote the raft's length and width, respectively, then its volume V is given by
V = 18 L W
Differentiating both sides with respect to time t gives
dV/dt = (18 in) W dL/dt + (18 in) L dW/dt
We're told that dL/dt = -1 in/s and dW/dt = -2 in/s, so that at the moment L = 11 in and W = 4 in, the volume is changing at a rate of
dV/dt = (18 in) (4 in) (-1 in/s) + (18 in) (11 in) (-2 in/s)
dV/dt = -468 in³/s
so the raft is losing air at a rate of 468 in³/s. (Note that "losing" captures the negative sign from before.)