Answer
Magnitude: $30.9\,m/s^{2}$
Direction: $27.2^{\circ}$ above the positive x-axis.
Work Step by Step
According to Newton's second law, $F_{net}=ma$
$\implies a=\frac{F_{net}}{m}$
Let A and B be the magnitudes of the two forces acting on the object and $\theta$ be the angle between them. Then the magnitude of the resultant force is
$F_{net}=\sqrt {A^{2}+B^{2}+2AB\cos\theta}
=\sqrt {(40.0\,N)^{2}+(60.0\,N)^{2}+2\times40.0\,N\times60.0\,N\times\cos45^{\circ}}$
$=92.7\,N$
Let $\alpha$ be the angle the resultant force makes with the positive x-axis.
Then $\tan \alpha=\frac{B\sin\theta}{A+B\cos\theta}=\frac{60.0\,N\times\sin45^{\circ}}{40.0\,N+60.0\,N\times\cos45^{\circ}}=0.514$
Or $\alpha=\tan^{-1}(0.514)=27.2^{\circ}$
Now, the magnitude of the acceleration is $a=\frac{F_{net}}{m}=\frac{92.7\,N}{3.00\,kg}=30.9\,m/s^{2}$
The direction of acceleration is the same as the direction of the force. It is directed $27.2^{\circ}$ above the positive x-axis.
