A piano wire weighing 6.00 g and having a length of 90.0 cm...

Mass of the wire, $m = 6.00 \text{ g} = 6 \times 10^{-3} \text{ kg}$

Length of the wire, $L = 90.0 \text{ cm} = 0.9 \text{ m}$

Fundamental frequency, $\nu = 261.63 \text{ Hz}$

The fundamental frequency of a stretched string is given by:

$\nu = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}$

where $\mu = \dfrac{m}{L}$ is the linear mass density.

Squaring both sides and rearranging for tension $T$:

$T = 4 L^2 \nu^2 \mu = 4 L m \nu^2$

Substituting the given values:

$T = 4 \times 0.9 \times 6 \times 10^{-3} \times (261.63)^2$

$T = 0.0216 \times 68450.26$

$T \approx 1478.5 \text{ N}$

Rounding off to three significant figures, we get:

$T = 1480 \text{ N}$