A

Step 1: The energy needed to polarize a dipole in an electric field is given by the potential energy formula: \[ U = -\vec{p} \cdot \vec{E} \] where \( \vec{p} \) is the dipole moment and \( \vec{E} \) is the electric field. We can express this as: \[ U = -pE \cos(\theta) \] where \(\theta\) is the angle between the dipole moment and the electric field.

Step 2: Before the field direction changes, the energy is: \[ U_1 = -pE \cos(0) = -pE \]
Inserting the provided values: \( p = 10^{-29} \, \text{C m}, \, E = 10^6 \, ext{V/m} \):
\[ U_1 = -(10^{-29}) (10^6) = -10^{-23} \, ext{J} \]

Step 3: After the field direction changes to 60º, the energy becomes: \[ U_2 = -pE \cos(60º) = -pE \cdot \frac{1}{2} \]
Now calculate \( U_2 \):
\[ U_2 = -(10^{-29}) (10^6) \cdot \frac{1}{2} = -0.5 \times 10^{-23} \, ext{J} \]

Step 4: Now, calculate the heat released, which is the difference in energy: \[ Q = U_2 - U_1 = (-0.5 \times 10^{-23}) - (-10^{-23}) \]
Simplifying this: \[ Q = 0.5 \times 10^{-23} \, ext{J} \]

Step 5: Therefore, the heat released by the substance in aligning its dipoles along the new direction of the field is \( 0.5 \times 10^{-23} \, ext{J} \).
The answer fits into option A as a significant approximation.