ANSWER
![3\pi sq.\: in.](https://img.qammunity.org/2020/formulas/mathematics/high-school/u9mlammqbp0vaqyszz28xry3ymnmt8d878.png)
EXPLANATION
Let R be the radius of the bigger circle and r, be the radius of the smaller circle.
Then their ratio is given as,
![R:r=3:1](https://img.qammunity.org/2020/formulas/mathematics/high-school/a0071qppwm9ub5truu47kcr4416lbzm1fp.png)
We can rewrite it as fractions to get,
![(R)/(r) = (3)/(1)](https://img.qammunity.org/2020/formulas/mathematics/high-school/zo62nv5cmzo1rqvvg1zwjz65416g62pv13.png)
We make R the subject to get,
![R = 3r](https://img.qammunity.org/2020/formulas/mathematics/high-school/gaa6h6s4k9haveh22gqx71ozj7i4fk9irs.png)
The area of the bigger circle can be found using the formula,
![Area=\pi {r}^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/vhnhyt9m0a8g6azd2nj3wvxt4ngh4g6fzt.png)
This implies that,
![Area=\pi ({3r})^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/a0dmta36pxu6boavs5ppmazl09gm2tc57o.png)
![Area=9\pi {r}^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/6qbbntxle7wzlbgczcri2dtboz95ug7fi5.png)
But it was given in the question that, the area of the bigger circle is 27π.
![27\pi=9\pi {r}^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/g3fow849wk9s6ekg0nqwr4htt9lr8igjwy.png)
We divide through by 9π to get,
![3 = {r}^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/hr00pgtcmzo80cwyan9u4k8nz5ae1h4kog.png)
This means that,
![r = √(3)](https://img.qammunity.org/2020/formulas/mathematics/high-school/8af9ixirr6u8cvml3q936gijiw8a0ohsml.png)
The area of the smaller circle is therefore
![= \pi {( √(3)) }^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/w85grvb21yyxd9it1bqgw329xxk570sgac.png)
![= 3\pi](https://img.qammunity.org/2020/formulas/mathematics/high-school/6sh6azf02svkj12nd26u5z30vwnwoz5thf.png)