Answer
1.4 $m/s^{2}$ in the direction opposite to that of the female skater's acceleration.
Work Step by Step
Let $\vec{F_{1}}$ be the force the male skater applies on the female skater and $\vec{F_{2}}$ be the force the female skater exerts on the male skater. Then, according to Newton's third law,
$\vec{F_{1}}=-\vec{F_{2}}$ (negative sign shows that the direction is opposite.)
As magnitudes are equal, we can write
$F_{1}=F_{2}$ or $m_{1}a_{1}=m_{2}a_{2}$
$\implies a_{1}=\frac{m_{2}a_{2}}{m_{1}}=\frac{45\,kg\times2.0\,m/s^{2}}{65\,kg}=1.4\,m/s^{2}$
