Given,

For first order reflection, 

$Ø_1 = 3.4°$

To find, 

$Ø_2 = ??$

Sice it is the same monochromatic x-ray, values of d(interstitial distance) and L(here lambda)

Using Bragg's Law,  $2d\sin\phi = nL$

For first order reflection,

$2d\sin(3.4) = (1)L$......(1)

For second order reflection,

$2d\sinØ_2 = (2)L$ .......(2)

Dividing (1) and (2),

=>  $\dfrac{\sin3.4}{\sinØ_2} = \dfrac{1}{2}$

=>  $\sin Ø_2 = 2 \times \sin3.4$

=> Ø2 = 6.25°