Answer
$3.20cm$
Work Step by Step
We can find the distance between the two second order minima as follows:
$\frac{\lambda}{W}=\frac{sin\theta}{m}$
$\implies \frac{\lambda}{W}=\frac{sin(0.458^{\circ})}{1}=0.008$
For the second diffraction minima
$sin\theta_2=\frac{m\lambda}{W}$
$\implies sin\theta_2=2(\frac{\lambda}{W})$
$\implies sin\theta_2=2\times 0.008=0.016$
$\implies \theta_2=sin^{-1}(0.0016)=0.916^{\circ}$
We know that
$y_2=Ltan\theta_2$
$\implies y_2=1.0m tan(0.916^{\circ})=0.016m$
Now the distance between the two second order minima is given as
$2y_2=2\times 0.016m=0.032m=3.20cm$
