Find the required annual interest rate to the nearest tenth of a percent for $\$ 1400$ to grow to $\$ 1800$ if interest is compounded quarterly for $6 \mathrm{yr}$.

Step 1 ：We are given that the principal amount (P) is $1400, the final amount (A) is $1800, the interest is compounded quarterly so n = 4, and the time (t) is 6 years.

Step 2 ：We can use the formula for compound interest, which is A = P (1 + r/n)^(nt), to find the annual interest rate (r).

Step 3 ：We can rearrange the formula to solve for r: r = n[(A/P)^(1/nt) - 1].

Step 4 ：Substitute the given values into the formula: r = 4[(1800/1400)^(1/(4*6)) - 1].

Step 5 ：Solving the equation gives r = 0.04210580740480552.

Step 6 ：Convert r to a percentage to get the annual interest rate: r_percent = 4.210580740480552.

Step 7 ：Round r_percent to the nearest tenth of a percent to get the final answer.

Step 8 ：Final Answer: The required annual interest rate to the nearest tenth of a percent for $1400 to grow to $1800 if interest is compounded quarterly for 6 years is \(\boxed{4.2\%}\).