1. For small oscillations of a magnetic needle in a uniform magnetic field,

$T = 2\pi \sqrt{\frac{I}{MB}}$

where:

• $T$ = time period
• $I$ = moment of inertia
• $M$ = magnetic moment
• $B$ = magnetic field

2. Rearranging for $B$:

$B = \frac{4\pi^2 I}{M T^2}$

3. Given data:

$M = 9.85 \times 10^{-2}\ \text{A m}^2$ $I = 5 \times 10^{-6}\ \text{kg m}^2$

The needle makes 10 oscillations in 5 s, so

$T = \frac{5}{10} = 0.5\ \text{s}$

Also given:

$\pi^2 = 9.85$

4. Substitute into the formula:

$B = \frac{4 \times 9.85 \times 5 \times 10^{-6}}{(9.85 \times 10^{-2})(0.5)^2}$

5. Simplify numerator:

$4 \times 9.85 \times 5 \times 10^{-6} = 197 \times 10^{-6}$

Denominator:

$(9.85 \times 10^{-2})(0.5)^2 = 9.85 \times 10^{-2} \times 0.25$ $= 2.4625 \times 10^{-2}$

Thus,

$B = \frac{197 \times 10^{-6}}{2.4625 \times 10^{-2}}$

Since $197/2.4625 = 80$,

$B = 80 \times 10^{-4}\ \text{T} = 8 \times 10^{-3}\ \text{T}$

6. Convert to mT:

$8 \times 10^{-3}\ \text{T} = 8\ \text{mT}$

Therefore, the required magnetic field is

$\boxed{8}$

mT.