Solution:

The area vector \( \vec{A} \) for a surface in the \( YZ \)-plane points along the \( \hat{i} \)-direction, with magnitude \( A = 20 \). Thus,
\[
\vec{A} = 20\hat{i}.
\]

The electric field is given by:
\[
\vec{E} = 5\hat{i} + 4\hat{j} + 9\hat{k}.
\]

Flux \( \Phi \) is:
\[
\Phi = \vec{E} \cdot \vec{A} = (5\hat{i} + 4\hat{j} + 9\hat{k}) \cdot 20\hat{i}.
\]

Only the \( \hat{i} \)-component contributes:
\[
\Phi = 5 \times 20 = 100 \, \text{units}.
\]