The diffraction pattern of a crystalline solid gave a peak at $2 \theta = 60^{\circ}$. What is the distance (in $cm$) between the layers which gave this peak? ($\lambda$ of $X$-rays is $1.54 \mathring{A}$) ($\sin 30^{\circ} = 0.5$,$\sin 60^{\circ} = 0.866$; $n = 1$)

• A
  $8.89 \times 10^{-9}$
• B
  $8.89 \times 10^{-1}$
• C
  $1.54 \times 10^{-8}$
• D
  $1.54$