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.
