You can find the general technique for balancing redox equations in acid solution here.
We know that #"IO"_3^-# is reduced to #"I"^-# and #"Ag"# is oxidized to #"Ag"^+#.
We also know that #"Ag"^+"(aq)" + "I"^(-)"(aq)" → "AgI(s)"#, so the iodide will exist as a precipitate of silver iodide.
Step 1: Write the two half-reactions.
#"IO"_3^(-) → "AgI"#
#"Ag" → "Ag"^+#
Step 2: Balance all atoms other than #"H"# and #"O"#.
#"IO"_3^(-) + "Ag"^+ → "AgI"#
#"Ag" → "Ag"^+#
Step 3: Balance #"O"#.
#"IO"_3^(-) + "Ag"^+ → "AgI" + color(red)(3)"H"_2"O"#
#"Ag" → "Ag"^+#
Step 4: Balance #"H"#.
#"IO"_3^(-) + "Ag"^+ + color(blue)(6)"H"^+ → "AgI" + color(red)(3)"H"_2"O"#
#"Ag" → "Ag"^+#
Step 5: Balance charge.
#"IO"_3^(-) + "Ag"^+ + color(blue)(6)"H"^+ + color(green)(6)"e"^(-)→ "AgI" + color(red)(3)"H"_2"O"#
#"Ag" → "Ag"^+ + color(magenta)(1)"e"^(-)#
Step 6: Equalize electrons transferred.
#color(magenta)(1) × ["IO"_3^(-) + "Ag"^+ + color(blue)(6)"H"^+ + color(green)(6)"e"^(-)→ "AgI" + color(red)(3)"H"_2"O"]#
#color(green)(6) × ["Ag" → "Ag"^+ + color(magenta)(1)"e"^(-)]#
Step 7: Add the two half-reactions.
#"IO"_3^(-) + color(red)(cancel(color(black)("Ag"^+))) + color(blue)(6)"H"^+ + color(red)(cancel(color(black)(color(green)(6)"e"^(-)))) → "AgI" + color(red)(3)"H"_2"O"#
#6"Ag" → stackrel(5)(color(red)(cancel(color(black)(6))))"Ag"^+ + color(red)(cancel(color(black)(color(magenta)(6)"e"^(-))))#
#stackrel(———————————————————)("IO"_3^(-) + 6"Ag" + color(blue)(6)"H"^+ → "AgI"+ 5"Ag"^+ + color(red)(3)"H"_2"O")#
Step 8: Check mass balance.
On the left: #"1 I; 3 O; 6 Ag; 6 H"#
On the right: #"6 Ag; 1 I; 6 H; 3 O"#
Step 9: Check charge balance.
On the left: #"1- + 6+ = 5+"#
On the right: #"5+"#
∴ The balanced equation is
#"IO"_3^(-) + 6"Ag" + 6"H"^+ → "AgI"+ 5"Ag"^+ + 3"H"_2"O"#
