#### Answer

The plane is traveling at 1.41 times the speed of sound (aka Mach 1.41).

#### Work Step by Step

In the same time that it takes sound to travel from point A to point C, the plane has moved from point A to point B.
Triangle ACB is a special 45-45-90 right triangle. The distance traveled by the plane, AB, is $\sqrt{2} \times AC$.
An internet search for the term "Mach angle" shows that generally, the sine of the half-angle gives the ratio of the speed of sound to the speed of the plane.
$$sin(45^{\circ}) = \frac{1}{\sqrt{2} } = \frac{v_{sound}}{v_{plane}}$$
This gives the same answer.