Given wave equation

The transverse wave is $y=-0.21\sin(x+30t)$

Compare with the standard form: $y=A\sin(kx\pm \omega t)$

Hence, $k=1\ \text{rad/m},\qquad \omega=30\ \text{rad/s}$

Find wave speed

Wave speed is $v=\frac{\omega}{k}=\frac{30}{1}=30\ \text{m/s}$

Convert linear mass density to SI units

Given: $\mu=0.135\ \text{g/cm}$

Now, $1\ \text{g/cm}=\frac{10^{-3}\ \text{kg}}{10^{-2}\ \text{m}}=10^{-1}\ \text{kg/m}$

Therefore, $0.135\ \text{g/cm}=0.135\times 10^{-1}=0.0135\ \text{kg/m}$

Use the relation for wave speed on a stretched string

$v=\sqrt{\frac{T}{\mu}}$

So, $T=\mu v^2$

Substituting, $T=0.0135\times (30)^2$

$T=0.0135\times 900=12.15\ \text{N}$

Match with the required form

The tension is written as $T=x\times 10^{-2}\ \text{N}$

Since $12.15\ \text{N}=1215\times 10^{-2}\ \text{N}$

we get $x=1215$

Rounded to nearest integer

$x=1215$