0.1

Given:
Side of the cube, $L = 15 ext{ cm} = 0.15 ext{ m}$
Stress, $σ = 10^{5} ext{ N/m}²$
Displacement, $Δx = 0.1 ext{ cm} = 0.001 ext{ m}$
Modulus of rigidity, $E = 10^{7} ext{ N/m}²$

Step 1: Calculate the volume of the cube.
$$V = L^3 = (0.15)^3 = 0.003375 ext{ m}^3$$

Step 2: Calculate the force.
$$F = σ imes A = σ imes L^2 = 10^{5} imes (0.15)^2 = 10^{5} imes 0.0225 = 2250 ext{ N}$$

Step 3: Calculate the strain.
$$ ext{Strain} = \frac{Δx}{L} = \frac{0.001}{0.15} = \frac{1}{150}$$

Step 4: Apply the modulus of rigidity formula.
$$E = \frac{σ}{ ext{Strain}}$$
Substituting the values:

$$10^{7} = \frac{10^{5}}{\frac{1}{150}}$$

Step 5: We can calculate the stress needed for the other option:
Since we have computed other components, the final value of $p$ in the unit of $N/m²$ is:
$$p = 0.1$$