1. The problem is to find the value of $\sqrt{243}$.\n\n2. First, factorize 243 into its prime factors.\n$$243 = 3 \times 81 = 3 \times 3 \times 27 = 3 \times 3 \times 3 \times 9 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5$$\n\n3. Use the property of square roots that $\sqrt{a^2} = a$. So,\n$$\sqrt{243} = \sqrt{3^5} = \sqrt{3^4 \times 3} = \sqrt{(3^2)^2 \times 3} = 3^2 \times \sqrt{3} = 9\sqrt{3}$$\n\n4. Therefore, the simplified form of $\sqrt{243}$ is $9\sqrt{3}$.