Answer
The electric field has a strength of $4.5\times 10^4~N/C$ directed toward the bead.
Work Step by Step
We can find the magnitude of the electric field 4.0 cm from the bead:
$E = \frac{k~q}{r^2}$
$E = \frac{(9.0\times 10^9~N~m^2/C^2)(8.0\times 10^{-9}~C)}{(0.040~m)^2}$
$E = 4.5\times 10^4~N/C$
An electric field points toward a negative charge. Therefore, the electric field has a strength of $4.5\times 10^4~N/C$ directed toward the bead.
