const regex = /(\s*["]\s*)/gm;

var test_string =  = `{" 1 ": "b.) some pasta salad", "   10": "   a.) Yes, some bread", "11  ": "   a.) eggs and toast  " }
`;

console.log(test_string.replace(regex, '"'))

假设字符串内容始终是有效的JSON，您可以这样做：

var str = '{" 1 ": "b.) some pasta salad", "   10": "   a.) Yes, some bread", "11  ": "   a.) eggs and toast  " }'

const res=JSON.stringify(Object.fromEntries(Object.entries(JSON.parse(str)).map(e=>e.map(e=>e.trim()))))

console.log(res)

我知道，这是一种“愚蠢”的解决方案，但它适用于给定的示例字符串。
代码片段解析JSON字符串，然后将结果对象转换为数组的数组，每个子数组的每个元素都是.trim()-med，在操作结束时，通过.Object.fromEntries()重构对象，并通过JSON.stringify()将其转换回JSON字符串。

将零个或多个空格后接引号，再将零个或多个空格后接引号替换为引号：

replaceAll(/ *" */g, '"')

const test_string = '{" 1 ": "b.) some pasta salad", "   10": "   a.) Yes, some bread", "11  ": "   a.) eggs and toast  " }'

const trimmed = test_string.replaceAll(/ *" */g, '"');

console.log(trimmed);

const test_string = '{" 1 ": "b.) some pasta salad", "   10": "   a.) Yes, some bread", "11  ": "   a.) eggs and toast  " }';

const test_string_format = test_string.replace(/\s*"\s*/g, '"');

console.log(test_string_format);

• \s：匹配任何空白字符
• *：匹配零个或多个prev字符（\s）
• "：匹配双引号字符

console.log('{" 1 ": "b.) some pasta salad", "   10": "   a.) Yes, some bread", "11  ": "   a.) eggs and toast  " }'
    .replaceAll(/( *\" +)|( +\ *)/ig,'"'))